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11 July, 17:11

An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 * 10-19 c and its mass is 9.10939 * 10-31 kg. how strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.9 cm? answer in units of t.

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  1. 11 July, 17:18
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    By definition, the potential energy is:

    U = qV

    Where,

    q: load

    V: voltage.

    Then, the kinetic energy is:

    K = mv ^ 2/2

    Where,

    m: mass

    v: speed.

    As the power energy is converted into kinetic energy, we have then:

    U = K

    Equating equations:

    qV = mv ^ 2/2

    From here, we clear the speed:

    v = root (2qV / m)

    Substituting values we have:

    v = root ((2 * (1.60218 * 10 ^ - 19) * 3600) / 9.10939*10^-31))

    v = 3.56 * 10 ^ 7 m / s

    Then, the centripetal force is:

    Fc = Fm

    mv ^ 2 / r = qvB

    By clearing the magnetic field we have:

    B = mv / qr

    Substituting values:

    B = (9.10939 * 10 ^ - 31) * (3.56 * 10 ^ 7) / (1.60218 * 10 ^ - 19) * 0.059

    B = 3.43 * 10 ^ - 3 T

    Answer:

    A magnetic field that must be experienced by the electron is:

    B = 3.43 * 10 ^ - 3 T
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